1 Answer
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Note, as Didier Piau implies in the comments, your function is not a density. I believe you want $f(x_1,x_2)=8 x_1x_2$ for $0<x_1<x_2<1$ and $f(x_1,x_2)=0$, otherwise. Note that the density is $0$ below the line $\color{gray}{x_2=x_1}$.

To find the probability that $X_1+X_2\le1$, sketch the region first. It's the region beneath the line $\color{darkgreen}{x_2=1-x_1}$ (the pink and green regions in the diagram below). But since the density is 0 beneath the line $\color{gray}{x_2=x_1}$, we may just consider the pink region shaded below (over the green region, we're integrating to zero function).

So, to set up the integral over the pink region, let's think of that region as being generated by the line segments $\color{darkgreen}{\ell_{x_1}}$ that "sweep" across it from $x_1=0$ to $x_1=1/2$.
The inner integral corresponds to integrating over a fixed line segment, $\color{darkgreen}{\ell_{x_1}}$ , with respect to $x_2$. The lower limit is the bottom of the line segment and the upper limit is the top of the line segment. Make sure the limits of integration are in terms of $x_1$:
$$
\int_{x_1}^{1-{x_1}} f(x_1,x_2)\,dx_2.
$$
Then you integrate the above expression as the line segments range from $x_1=0$ to $x_1=1/2$:
$$
\int_0^{1/2}\int_{x_1}^{1-{x_1}} f(x_1,x_2)\,dx_2\, dx_1.
$$